Take the 2-minute tour ×
Mathematics Stack Exchange is a question and answer site for people studying math at any level and professionals in related fields. It's 100% free, no registration required.

Just before the truth-arrows in a topos subsection of Goldblatt's "Topoi: A Categorial Analysis of logic," descriptions of the truth functions $\Rightarrow$ and $\smallsmile$ are given in $\mathbf{Set}$ in terms of arrows. This is to motivate their abstraction to any given topos.

I'm having trouble understanding these descriptions thoroughly, so I would like some clarification, please :)

Definition 1: The disjunction truth-function $\smallsmile :2\times 2\to 2$ (for $2=\{0, 1\}$) is given by $1\smallsmile 1=0\smallsmile1=1\smallsmile0=1$ and $0\smallsmile0=0$, whereas the implication truth-function $\Rightarrow: 2\times 2\to 2$ is given by $1\Rightarrow0=0$ and $(0\Rightarrow1)=(0\Rightarrow0)=(1\Rightarrow1)=1$.

Definition 2: $\smallfrown$ can be taken as the characteristic function of the product map $\langle\operatorname{true}, \operatorname{true}\rangle$.


Implication.

Goldblatt states that $\Rightarrow$ is the characteristic function of enter image description here$=\{\langle 0, 0\rangle,\langle 0, 1\rangle, \langle 1, 1\rangle\}$ with

enter image description here

a pullback square, which he then uses to write enter image description here as the equaliser of $\smallfrown:2\times 2\to 2$ and $pr_1$, where $pr_1(\langle x,y\rangle)=x$. I don't see how he does this.


Disjunction:

We take $\smallsmile$ as $\chi_D$ for $D=A\cup B$ for $A=\{\langle1, 1\rangle, \langle1, 0\rangle\}$ and $B=\{\langle1, 1\rangle,\langle0, 1\rangle\}$; identify $A$ with the product map $\langle\operatorname{true}_2, 1_2\rangle$, $B$ with $\langle1_2, \operatorname{true}_2\rangle$; form the coproduct map $f=[\langle\operatorname{true}_2, 1_2\rangle, \langle1_2, \operatorname{true}_2\rangle]$; identify $\operatorname{im} f=D$; epic-monic factor $f$ through $D$; then claim that all this specifies $D$ - and hence $\small\smile$ - uniquely up to isomorphism.

So yeah, I can't see the wood for the trees basically. Please help :)

share|improve this question
    
Okay, things are starting to make sense . . . –  Shaun Jun 28 at 18:18
    
Ah, I get it. It's easy! I'm overthinking it. I'll type up my answer tomorrow :) –  Shaun Jun 28 at 21:29

1 Answer 1

up vote 2 down vote accepted

I was trying too hard.

Implication.

When Goldblatt used the lattice equivalence $$x\sqsubseteq y \iff x=x\sqcap y$$ to write enter image description here as $$\{\langle x, y\rangle\vert x\smallfrown y=x\},$$ he did so for a reason: it's so that enter image description here is the set for which $\smallfrown$ and $pr_1$ agree, which is then, by $\S 3.10$, the equaliser of the two.

Disjunction.

If you draw the commutative diagrams for the products $t_1=\langle\operatorname{true}_2=\operatorname{true}\circ\vert_2, 1_2\rangle$ and $t_2=\langle1_2,\operatorname{true}_2\rangle$ and write down all the equations (like $pr_2\circ t_1=1_2$), you'll see that $t_1$ and $t_2$ are nothing more than what Goldblatt said they were: $t_1$ sends $0$ to $\langle1, 0\rangle$ and $1$ to $\langle1, 1\rangle$; $t_2$, similarly; and we can identify them with $A$ and $B$, respectively, because $A$ and $B$ are two-element sets whose set-inclusions with respect to $2\times 2$ are $t_1$ and $t_2$, respectively.

The coproduct $2+2$ is just the disjoint union of $2$ with itself, so that we can write $$2+2=2^{(0)}\cup2^{(1)},$$ where $2^{(0)}=2\times\{0\}$ and $2^{(1)}=2\times\{1\}$, to see that $f$ is given by $$f(a, i)=\begin{cases}t_1(a) &: i=0 \\ t_{2}(a) &: i=1.\end{cases}$$

The rest is downhill from there (once you see that $\operatorname{im}f=\operatorname{im}t_1\cup\operatorname{im}t_2$).

share|improve this answer

Your Answer

 
discard

By posting your answer, you agree to the privacy policy and terms of service.

Not the answer you're looking for? Browse other questions tagged or ask your own question.